If `A = underset(n rarr oo)("Lt") sum_(r = 1)^(n) tan^(-1) ((2r)/(2 + r^(2) + r^(4)))` then A is equal to

Options

A. `pi//3`
B. `pi//4`
C. `pi//6`
D. none

Answer

To solve the problem, we need to evaluate the limit:

\[ A = \lim_{n \to \infty} \sum_{r=1}^{n} \tan^{-1} \left( \frac{2r}{2 + r^2 + r^4} \right) \]

### Step-by-Step Solution:

1. **Simplifying the Denominator**:
We start by simplifying the expression inside the arctangent. The denominator can be rewritten:
\[ 2 + r^2 + r^4 = r^4 + r^2 + 2 = (r^2 + 1)^2 - r^2 \]
This can be factored as:
\[ (r^2 + 1 - r)(r^2 + 1 + r) \]

2. **Rewriting the Arctangent**:
We can express the arctangent in a different form:
\[ \tan^{-1} \left( \frac{2r}{(r^2 + 1 - r)(r^2 + 1 + r)} \right) \]
This can be viewed as:
\[ \tan^{-1}(x) - \tan^{-1}(y) \text{ where } x = r^2 + r + 1 \text{ and } y = r^2 - r + 1 \]
Thus, we have:
\[ \tan^{-1}(r^2 + r + 1) - \tan^{-1}(r^2 - r + 1) \]

3. **Summing the Series**:
Now, we can rewrite the sum:
\[ \sum_{r=1}^{n} \left( \tan^{-1}(r^2 + r + 1) - \tan^{-1}(r^2 - r + 1) \right) \]
This is a telescoping series. Most terms will cancel out:
\[ = \tan^{-1}(n^2 + n + 1) - \tan^{-1}(1^2 - 1 + 1) \]
The first term approaches \(\tan^{-1}(n^2 + n + 1)\) and the last term is \(\tan^{-1}(1)\).

4. **Taking the Limit**:
As \(n\) approaches infinity:
\[ \tan^{-1}(n^2 + n + 1) \to \frac{\pi}{2} \]
Therefore:
\[ A = \lim_{n \to \infty} \left( \tan^{-1}(n^2 + n + 1) - \tan^{-1}(1) \right) = \frac{\pi}{2} - \frac{\pi}{4} \]

5. **Final Calculation**:
Thus:
\[ A = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Conclusion:
The value of \(A\) is:

\[ \boxed{\frac{\pi}{4}} \]